Post on 19-Jan-2016
In-span loads on beam elements
So far we have only been able to apply loads at nodes.
example
How do we then tackle loads away from nodes, or continuous loads?---------------------------------------------------------------Option 1: add a node.
example
Option 2: fool the structure into thinking it has in-span loads when it doesn’t. This is the (more powerful) technique we will study in detail.
We have to apply loads/moments at nodes which have the same effect on the structure as the in-span loads.
We conduct two analyses
(A)Clamp all the nodes in the structure, apply the in-span loads and work out the reactions at the clamps.
(B)Release the clamps, remove the in-span loads and reverse the reactions at nodes and conduct a standard matrix stiffness method analysis of the structure.
Why does this work?
It is the principle of superposition which relates to linear elastic structures only.
It does not matter in which order you apply loads to a structure, the deflections/rotations will be the same
An example – a propped cantilever
P Deflected shape under load ( this is what we want to find out)
P
Analysis A
Analysis B
Fixed end moments and
forces
So that we do not ever have to carry out analysis A we use tables of “answers” from A type analyses.
Example – why can we get away with this?
Fixed end moments/reactions are available for a number of load cases (tables in many textbooks) one is on DUO.
Can you derive them? (Yes).
udl of w/m
L
Properties: E, I, A
An example – cantilever with a UDL
We know that:End deflection =
End slope =
Vertical support reaction =
Moment support reaction =
EI
wL
8
4
EI
wL
6
3
wL
2
2wL
1 21
1
1
V
U
2
2
2
V
U1
1
1
M
F
F
Y
X
2
2
2
M
F
F
Y
X
2
2
2
23
2 46
612
12
2V
L
EI
L
EIL
EI
L
EI
wL
wL
We can ignore axial effects as well as those d.o.f.s which are fixed
P
udl of w/m
2L
L2L/3
A
B C
D
Node nos.
PFE effects example 2: Portal frame
Properties
E,I, A
45o
2
1
3
4
P
udl o f w /m
2 L
L2 L /3
A
B C
D
N ode nos.
PPortal frame example
“first load” vector, i.e. those loads already at nodes
2
1
3
4
00
00 0 01XF 1YF 1M 4XF 4YF 4M2P
2P
Nodal load vector (transposed)
Analysis A
27
4PL
27
2PL
27
20P
27
7P
wL wL
3
2wL
3
2wL
wL wL
3
2wL
3
2wL
27
4PL
27
2PL
27
20P
27
7P
Node 3
Node 4Node 1
Node 2
FE effects force vector
27
2PL 27
7PNode 1
00
FE effects force vector
27
7P0
27
2PL
wL
3
2wL
27
4PL
27
20P
Node 2
00
27
7P0
27
2PL27
20PwL
327
4 2wLPL
wL
3
2wL
Node 3
Node 4
00
27
7P0
27
2PL27
20PwL
327
4 2wLPL
wL3
2wL0
Node 4
00
27
7P0
27
2PL27
20PwL
327
4 2wLPL
wL3
2wL0 0 0 0
00
27
7P0
27
2PL27
20PwL
327
4 2wLPL
wL3
2wL0 0 0 0
Bringing the two load vectors together
“first load” vector, i.e. those loads already at nodes
00
00 0 01XF 1YF 1M 4XF 4YF 4M2P
2P
So what we now have to solve is this
4
4
4
3
3
3
2
2
2
1
1
1
4
4
4
2
2
1
1
1
matrix Stiffness Global1212
3
2
2
327
4
2720
272
277
V
U
V
U
V
U
V
U
M
F
F
wLwLP
P
wLPLwL
P
PLM
F
PF
Y
X
Y
X
FE effects example 3: 2 span beam
Error lurking
udl of w/m
L
Results of stiffness matrix
analysis
"Free" structure
"Free" BMD
Final BMD
This is a propped cantilever
In-span bending moments